When you catch a ball of mass
m that was travelling at velocity v the change of momentum of the ball during the catching is mv.
To do this you must apply a force F for a time t. So since impulse equals change of momentum Ft
= mv. You can see that the longer you take to bring the ball to rest the smaller the force - that's
why it is sensible to "bring your hands back" when catching a ball - especially a hard one like a
cricket ball, the ball does not "give" so you must.

When you hit the ball straight
back the change of momentum is much bigger.

Momentum of ball before collision =
mv

Momentum of ball after collision = - mv [momentum is a vector and we must allow for the
change of direction, the minus sign tells you that the ball is moving in the opposite direction].

Therefore change of momentum of ball = mv - (-mv) = 2mv

So force on the ball
F = 2mv/t and if the time spent in hitting the ball backwards is the same as that spent in catching
you can see that the force will be twice as large.

Catching :- Momentum change = mv Force = mv/t

Hitting back at the same speed :- Momentum change = 2mv Force = 2mv/t

Hitting back at the same speed :- Momentum change = 2mv Force = 2mv/t

1. A girl plays in a hockey match. At one point she stops a ball from a corner with her stick and at another she hits a ball that was coming towards her straight back. In both cases assume that the mass of the hockey ball was 0.15 kg, the velocity of the ball was 12 ms

(a) Stopping the ball: F = mv/t = 0.15x12/0.05 = 36 N

(b) Hitting the ball back: F = 2mv/t = 2x0.15x12/0.05 = 72 N

2. Emission of an alpha particle

A radium nucleus with a relative atomic mass of 226 (226Ra) emits an alpha particle (relative atomic mass 4 (

What is the recoil velocity of the radon nucleus?

Using: recoil velocity (V) = -vm/M = - 1.5x107x4/226 = - 2.65x105 ms